What is the mean free path of molecules, in nanometer, in an ideal gas in which the mean collision time is 5 × 10-10 s, the temperature is 374 K, and the mass of a molecule is 7 × 10-25 kg? Assume that the molecules are moving at their root-mean-square speeds. The Boltzmann constant is 1.38 × 10-23 J/K.
Question
What is the mean free path of molecules, in nanometer, in an ideal gas in which the mean collision time is 5 × 10-10 s, the temperature is 374 K, and the mass of a molecule is 7 × 10-25 kg? Assume that the molecules are moving at their root-mean-square speeds. The Boltzmann constant is 1.38 × 10-23 J/K.
Solution
The mean free path of molecules in an ideal gas can be calculated using the formula:
λ = v_rms * τ
where:
- λ is the mean free path,
- v_rms is the root-mean-square speed of the molecules, and
- τ is the mean collision time.
The root-mean-square speed can be calculated using the formula:
v_rms = sqrt(3kT/m)
where:
- k is the Boltzmann constant,
- T is the temperature, and
- m is the mass of a molecule.
Substituting the given values:
v_rms = sqrt(3 * 1.38 × 10^-23 J/K * 374 K / 7 × 10^-25 kg) v_rms = sqrt(1.5184 × 10^2 m^2/s^2) v_rms = 1.232 × 10^1 m/s
Substituting v_rms and τ into the formula for λ:
λ = 1.232 × 10^1 m/s * 5 × 10^-10 s λ = 6.16 × 10^-9 m
Converting meters to nanometers (1 m = 10^9 nm):
λ = 6.16 nm
So, the mean free path of the molecules in the ideal gas is approximately 6.16 nm.
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